The work done by an external force to move a -7.75 µC charge from point a to point b is 25.0 * 10-4 J. If the charge was started from rest and had 4.73*10-4 J of kinetic energy when it reached point b, what is the magnitude of the potential difference between a and b?
The charge times the potential difference is the increase in potential energy. That plus the acquired kinetic energy equals the work done.
(Vb - Va)*Q + 4.73*10^-4 J
= (25.0)*10^-4 J
I don't understand how you get the answer. What is the Vb, Va and Q?
To find the magnitude of the potential difference between points a and b, we can subtract the kinetic energy at point b from the work done by the external force.
Given that the work done by the external force is 25.0 * 10^-4 J and the kinetic energy at point b is 4.73 * 10^-4 J, we can calculate the potential difference as follows:
Potential difference = Work done - Kinetic energy at point b
Potential difference = (25.0 * 10^-4 J) - (4.73 * 10^-4 J)
Potential difference = 20.27 * 10^-4 J
Therefore, the magnitude of the potential difference between points a and b is 20.27 * 10^-4 J.
To find the magnitude of the potential difference between points A and B, we need to use the concept of work done and the equation for potential difference.
A charge moving in an electric field experiences a force that can do work on it. The work done by the external force to move the charge is equal to the change in its electrical potential energy.
The work done (W) on a charge (q) moving through a potential difference (V) can be calculated using the equation:
W = q * V
Given that the work done is 25.0 * 10^(-4) J and the charge is -7.75 µC, we can rearrange the equation to solve for the potential difference (V):
V = W / q
Substituting the values, we have:
V = (25.0 * 10^(-4) J) / (-7.75 *10^(-6) C)
V = -3.23 V
The potential difference between points A and B is -3.23 V. The negative sign indicates that the movement of the charge is opposite to the direction of the electric field.
Please note that the equation for kinetic energy is not required to solve this particular problem.